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March 15th, 2016, 04:40 PM   #11
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Quote:
Originally Posted by zylo View Post
I am discussing Cantor's Diagonal argument using his language.

Cantor's Diagonal Argument only works if the enumeration is wider than it is long.*

*Ref:
https://en.wikipedia.org/wiki/Cantor...gonal_argument
The article you've given as a reference doesn't mention that the enumeration is (or has to be) wider than it is long. You're making a false assertion and pretending that it's justified by an article that neither needs the assertion nor mentions it.
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March 15th, 2016, 04:46 PM   #12
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The reason the article doesn't mention it is because the enumerated sequences are of infinite length. The list cannot be longer than that.
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March 15th, 2016, 05:12 PM   #13
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Originally Posted by zylo View Post
For a finite number of rows, it is obvious.
For an infinite number of rows, you have to prove it.
You don't have to prove it when the sequences and rows are both infinite. You must already be aware of that, as your previous post states "There is no number infinity. You can't treat infinity as if it were a finite number."
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March 17th, 2016, 08:25 PM   #14
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Originally Posted by skipjack View Post
The article you've given as a reference doesn't mention that the enumeration is (or has to be) wider than it is long. You're making a false assertion and pretending that it's justified by an article that neither needs the assertion nor mentions it.
If it is longer than wide, Cantors argument doesn't work: post #4
If it is wider than long, Cantors argument doesn't work: post #7

Making statements about an infinite list of elements infinitely long is meaningless.
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March 17th, 2016, 08:45 PM   #15
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Zylo. The existence of at least one infinite set is an axiom of set theory. You are welcome to deny that axiom, but if you do you are working in a different system to the one that Cantor was working in.

If you deny the existence of infinite sets, Cantor is not wrong, he is simply irrelevant because if infinite sets do not exist, you obviously can't have a theorem about countably infinite sets of infinite series.

If you don't deny the existence of infinite sets, then infinite lists and infinite sequences come with the basic axioms. Again, you can't complain about Cantor's use of them.

What is meaningless is your persistent talking of rubbish about things you simply don't understand.
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March 17th, 2016, 11:32 PM   #16
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Quote:
Originally Posted by zylo View Post
If it is longer than wide, Cantor's argument doesn't work: post #4
That's not what you stated in post #4. You stated "Cantor's Diagonal Argument only works if the enumeration is wider than it is long." That assertion is untrue, as Cantor's diagonal argument works when the length of each sequence is a finite constant that equals the number of sequences. Also, the wikipedia article you gave as a reference for your false assertion doesn't make that assertion or any similar assertion.

Quote:
Originally Posted by zylo View Post
If it is wider than long, Cantor's argument doesn't work: post #7
That's obviously untrue if the length and width are finite. Your justification doesn't make sense in that case. That observation is sufficient to show that your post #7 is wrong.

Quote:
Originally Posted by zylo View Post
Making statements about an infinite list of elements infinitely long is meaningless.
You give no reason for this statement. It's "circular" anyway, so it's an invalid assertion by one of your own principles.
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March 17th, 2016, 11:58 PM   #17
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From post #4 (longer than wide)

1 0 1
0 1 0
1 0 0
1 1 1
0 0 1
. . . .
s = 0 0 1 is in the list

From post#7 Wider than long)

1) 0 1 0 0 1 1 0..............
.) 1 1 0 1 0 0 1..............
n) 0 0 1 0 1 0 1..............

n+1) 1 0 0 1 0 0 1..............

No matter what n is, Cantors sequence n+1 is in the list.

Cantor's diagonal argument doesn't lead to a contradiction.
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March 18th, 2016, 12:10 AM   #18
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In ZF, the "axiom of infinity" defines infinity as an inductive set.

https://en.wikipedia.org/wiki/Axiom_of_infinity
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March 18th, 2016, 12:27 AM   #19
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Quote:
Originally Posted by zylo View Post
From post #4 (longer than wide)
That's not what you gave in post #4. You gave "Cantor's Diagonal Argument only works if the enumeration is wider than it is long."
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March 18th, 2016, 01:19 AM   #20
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No matter what n is, Cantor's sequence n+1 is in the list.
You given no reason for that assertion, whereas Cantor gives a straightforward reason why it isn't - its "diagonal" construction ensures it differs from the first sequence in the first place, the second sequence in the second place, the nth sequence in the nth place, the n+1th sequence in the n+1th place, etc., so that it can't match any sequence already listed.
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